Evaluate `lim_(yto0)(y^(2)+sin x)/(x^(2)+siny^(2)),` where `(x,y)to(0,0)` along the curve `x=y^(2)`.

Explanation

To evaluate the limit

\[ \lim_{(x,y) \to (0,0)} \frac{y^2 + \sin x}{x^2 + \sin y^2} \]

along the curve \(x = y^2\), we will substitute \(x\) with \(y^2\) in the expression.

### Step 1: Substitute \(x = y^2\)

Substituting \(x = y^2\) into the limit gives us:

\[ \lim_{y \to 0} \frac{y^2 + \sin(y^2)}{(y^2)^2 + \sin(y^2)} \]

### Step 2: Simplify the expression

Now, we can simplify the expression:

\[ = \lim_{y \to 0} \frac{y^2 + \sin(y^2)}{y^4 + \sin(y^2)} \]

### Step 3: Analyze the limit

As \(y \to 0\), we know that \(\sin(y^2) \approx y^2\) (using the small angle approximation \(\sin z \approx z\) when \(z\) is small). Therefore, we can approximate:

\[ \sin(y^2) \approx y^2 \]

Substituting this approximation into our limit gives:

\[ = \lim_{y \to 0} \frac{y^2 + y^2}{y^4 + y^2} = \lim_{y \to 0} \frac{2y^2}{y^4 + y^2} \]

### Step 4: Factor out \(y^2\)

We can factor \(y^2\) from the denominator:

\[ = \lim_{y \to 0} \frac{2y^2}{y^2(y^2 + 1)} = \lim_{y \to 0} \frac{2}{y^2 + 1} \]

### Step 5: Evaluate the limit

Now, as \(y \to 0\):

\[ = \frac{2}{0^2 + 1} = 2 \]

Thus, the limit evaluates to:

\[ \boxed{2} \]